Chapter 3, Problem 3B.6ST

Interpretation Introduction

Interpretation:

The entropy of vaporization of benzene at ${\text{25}}^{\text{o}}\text{C}$ has to be calculated.

Concept Information:

Change in entropy on heating at constant volume: the change in entropy when the temperature is changed from ${\text{T}}_i$ to ${\text{T}}_{\text{f}}$. The greater the heat capacity of substance, the change in entropy increases by increasing temperature. A high heat capacity implies that a lot of heat is required to produce a given change in temperature must be more powerful than for when the heat capacity is low, and the entropy increase is correspondingly high.

$\text{ΔS}\text{=}\text{C}\text{ln}\left(\frac{{\text{T}}_{\text{f}}}{{\text{T}}_{\text{i}}}\right)$

Where,

C is the heat capacity of a substance,

${\text{T}}_{\text{f}}$ is the final temperature of the system,

${\text{T}}_i$ is the initial temperature of the system.

Trouton’s rule: In the vaporization of liquid, the condensed (or less disordered) phase of liquid vaporizes into the gaseous phase, where the dispersed gas occupies approximately same volume whatever its identity.

Then, considering value of the change in entropy of such liquids is constant as $\text{85}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Trouton’s rule exception: For exceptional liquids having special kind of interactions like hydrogen bonding and metallic bonding result in the liquid being less disordered than a random jumble of molecules. There exists high value of entropy change.

Entropy of vaporization: The entropy of vaporization at the boiling temperature of a liquid is related to its enthalpy of vaporization at that temperature.

${\text{Δ}}_{\text{vap}}\text{S}\text{=}\frac{{\text{Δ}}_{\text{vap}}\text{H}\left({\text{T}}_{\text{b}}\right)}{{\text{T}}_{\text{b}}}$

Where,

${\text{Δ}}_{\text{vap}}\text{S}$ is change in entropy of vaporization,

${\text{Δ}}_{\text{vap}}\text{H}$ is change in enthalpy of vaporization,

${\text{T}}_{\text{b}}$ is the boiling temperature.

Entropy of phase transition: In order to calculate the entropy of phase transition at a temperature other than the transition temperature, it is important to do additional calculations are made by steps as follows,

1. 1. Calculate the entropy change for heating liquid Benzene from ${\text{25}}^{\text{o}}\text{C}\text{to}{\text{70}}^{\text{o}}\text{C}$ using change in entropy on heating formula results ${\text{ΔS}}_1$
2. 2. Calculate the entropy of transition at ${\text{70}}^{\text{o}}\text{C}$ temperature using entropy of vaporization results ${\text{Δ}}_{\text{vap}}{\text{S}}_{\text{2}}$
3. 3. Calculate the change in entropy for cooling the vapor from ${\text{70}}^{\text{o}}\text{C}\text{to}{\text{25}}^{\text{o}}\text{C}$ using change in entropy on heating formula results ${\text{ΔS}}_3$

On addition all the entropy contributions together gives the result as follows,

${\text{Δ}}_{\text{vap}}\text{S}\left(\text{298}\text{K}\right)\text{=}{\text{ΔS}}_{\text{1}}\text{+}{\text{Δ}}_{\text{vap}}{\text{S}}_{\text{2}}\text{+}{\text{ΔS}}_{\text{3}}$

Explanation of Solution

Given: ${\text{T}}_{\text{b}}\text{=}\text{353}\text{.2}\text{K}$; ${\text{Δ}}_{\text{vap}}\text{H}\left({\text{T}}_{\text{b}}\right)\text{=}\text{30}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$; ${\text{C}}_{\text{p,m}}\left(\text{l}\right)\text{=}\text{136}\text{.1}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

${\text{C}}_{\text{p,m}}\left(\text{g}\right)\text{=}\text{81}\text{.6}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

1. 1. Entropy of change on heating:

Given: ${\text{T}}_{\text{i}}\text{=}\text{25}\text{+}\text{273K}\text{=}\text{298 K}$, ${\text{T}}_{\text{f}}\text{=}\text{353}\text{.2 K}$ and ${\text{C}}_{\text{p,m}}\left(\text{l}\right)\text{=}\text{136}\text{.1}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

By substituting the values into the equation as follows,

$\begin{array}{c}{\text{ΔS}}_{\text{1}}\text{=}{\text{C}}_{\text{p,m}}\left(\text{benzene,}\text{l}\right)\text{ln}\left(\frac{{\text{T}}_{\text{f}}}{{\text{T}}_{\text{i}}}\right)\\ \\ \text{=}\left(\text{136}\text{.1 J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{ln}\left(\frac{\text{353}\text{.2}\text{K}}{\text{298}\text{K}}\right)\\ \\ \text{=}\left(\text{136}\text{.1 J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\left(\text{0}\text{.170}\right)\\ \\ \text{=}\text{23}\text{.137J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the change in entropy on heating is $23.137J.K^{-1}mo{l}^{-1}$

1. 2. Enthalpy of vaporization:

Given: ${\text{T}}_f\text{=}{\text{25}}^{\text{o}}\text{C}\text{+}\text{273K}\text{=}\text{298 K}$, ${\text{T}}_i\text{=}\text{353}\text{.2 K}$ and ${\text{C}}_{\text{p,m}}\left(\text{g}\right)\text{=}\text{81}\text{.6}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

$\begin{array}{c}{\text{Δ}}_{\text{vap}}{\text{S}}_{\text{2}}\text{=}\frac{{\text{Δ}}_{\text{vap}}\text{H}\left({\text{T}}_{\text{b}}\right)}{{\text{T}}_{\text{b}}}\\ \\ \text{=}\frac{\text{30}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}}{\text{353}\text{.2}\text{K}}\\ \\ \text{=}\text{87}\text{.2}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the estimated value of enthalpy of vaporization of benzene is $87.2J{K}^{-1}mo{l}^{-1}$

1. 3. Entropy of change on cooling:

Given: ${\text{T}}_{\text{i}}\text{=}{\text{25}}^{\text{o}}\text{C}\text{+}\text{273K}\text{=}\text{298 K}$ and ${\text{C}}_{\text{p,m}}\left(\text{l}\right)\text{=}\text{136}\text{.1}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

By substituting the values into the equation as follows,

$\begin{array}{c}{\text{ΔS}}_{\text{3}}\text{=}{\text{C}}_{\text{p,m}}\left(\text{benzene,}\text{g}\right)\text{ln}\left(\frac{{\text{T}}_{\text{f}}}{{\text{T}}_{\text{i}}}\right)\\ \\ \text{=}\left(\text{81}\text{.6 J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{ln}\left(\frac{\text{298}\text{K}}{\text{353}\text{.2}\text{K}}\right)\\ \\ \text{=}\left(\text{81}\text{.6 J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\left(\text{-}\text{0}\text{.170}\right)\\ \\ \text{=}\text{-13}\text{.87}\text{J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the change in entropy on cooling is $-13.87J.K^{-1}mo{l}^{-1}$

At last addition of all contribution leads to the result as follows,

$\begin{array}{l}{\text{Δ}}_{\text{vap}}\text{S}\left(\text{298}\text{K}\right)\text{=}{\text{ΔS}}_{\text{1}}\text{+}{\text{Δ}}_{\text{vap}}{\text{S}}_{\text{2}}\text{+}{\text{ΔS}}_{\text{3}}\\ \\ \text{=}\text{23}\text{.137}\text{J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+}\text{87}\text{.2}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+}\left(\text{-13}\text{.87}\text{J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ \\ \text{=}\text{96}\text{.467}\text{J}{\text{.K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

The sum of three entropy changes in entropy of transition at ${\text{25}}^{\text{o}}\text{C}$ is $96.4J.K^{-1}mo{l}^{-1}$